In the center of our milky way, it is assumed that a black hole exists with a mass of $\approx 4\times 10^6$ times our sun's mass. How much light bending (in degrees) would arise for stars that are in perspective near the position of this black hole in the sky?

Could the mass of a black holes become so great that it actually distorts the area in the sky we see around it (through telescopes)?

Right. For the Sun, the deflection angle was fixed and finite because the gravitational field - red shift - is finite at the Sun's surface. But the red shift may become arbitrarily extreme near the black hole. Light can even orbit a black hole - outside the event horizon.
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Luboš MotlFeb 21 '11 at 20:12

3 Answers
3

The general theory of gravitational lensing shows that a light ray which approaches within a radius $r~>>~2GM/c^2$ will be deflected approximately by an angle $\theta~=~GM/rc^2$. In a more general setting the deflection of light is given by the Einstein angular radius
$$
\theta_E~=~\sqrt{\frac{4GM}{c^2}\frac{d_{ls}}{d_ld_{s}}},
$$
where $d_{ls},~d_l,~d_s$ are the angular diameters to the gravitational lens, the source and the distance between the gravitational lens and the source. For $d_{ls},~d_l,~d_s$ the angular diameters to the gravitational lens, the source and the distance between the gravitational lens and the source. The condition $d_s~=~d_l~+~d_{sl}$ obtains locally where cosmological frame dragging is small. This theory is the weak gravitational lensing approximation, where the deflection of light is essentially a Newtonian result. The distance relationships are determined by $\theta d_s~=~\beta d_s~+~\alpha' d_{ls}$, for the angles given in the figure. The angle of deflection reduced angle of deflection $\alpha(\theta)~=~(d_{ls}/d_s)\alpha'(\theta)$ gives a relationship between the angles of importance $\alpha(\theta)~+~\beta~=~\theta$. The background which is then distorted scales as the square root of the mass of the black hole or large gravitating body. So at a fixed distance away the larger the steradian measure of distrotion is observed.

For the position of a source $\vec x$, the propagation of light along the $z$ axis from this source then reduces the visual appearance of the object to $\vec\xi~=~(\xi_x,~\xi_y)$ along the axis of optical propagation. The weak gravitational lensing of light then indicates that the deflection of the appearance of this object along the axis of optical propagation is given by the
$$
\Delta{\vec\xi}~=~\nabla\Phi(\xi),
$$
for $\xi$ the position of the image with the mass present and $\Phi(\xi)$ the gravitational potential. The difference in the vector position of the image ${\vec\xi}_i~-~{\vec\xi}_s$ is the difference between the position with the mass present and without it being present. The potential term obeys the Poisson equation so that
$$
\nabla^2\Phi~=~2\frac{\Sigma(\vec\xi)}{\Sigma_c}
$$
The integration over the direction of propagation then gives the mass density in the plane of the image, often called the surface mass density $\Sigma(\vec\xi)$. The angle of deflection $\alpha$ is then determined by the Poisson equation and the potential as
$$
{\vec\alpha}’(\vec\xi)~=~\frac{4G}{c^2}\int\frac{(\vec\xi~-~\vec\xi')\Sigma(\vec\xi')}{|\vec\xi~-~\vec\xi'|^2}d^2\xi',
$$
for $\Sigma(\vec\xi)$ a mass/area density distribution in the image. The function $\Sigma(\vec\xi)$ plays the role of an index of refraction based upon a mass distribution, which for a thin lens will give the angle of deviation. For a gravitational thin lens, a weak field that is very small compared to the optical path length, and $\Sigma(\vec\xi)$ is a constant. The deflection angle is simply
$$
\alpha(\xi)~=~\frac{4\pi G}{c^2}\frac{\Sigma(\xi) d_{ls}\xi}{d_s}
$$
where for small angles $|\vec\xi|~=\xi~=~d_l\theta$ and
$$
\alpha(\xi)~=~\frac{4\pi G\Sigma}{c^2}\frac{d_{ls}d_l}{d_s}~=~\frac{\Sigma}{\Sigma_c}\theta
$$
for the critical mass density $\Sigma_c~=~(c^2/4\pi G)(d_s/d_{ls}d_l)$. This is the minimal mass density which might be distributed in the area of an Einstein ring.

Famously, stars near the sun are deflected 1.75 arc seconds, as predicted by Einstein.
I calculate that stars on the other side of the galaxy whose light passed very near the galaxy center would be coincidentally be deflected by about the same amount.
If you observe in the submillimeter band you could see it, just as Doeleman sees the black hole itself.
But it would take far more than a lifetime for anything to pass behind the black hole at the center of the galaxy.
The apparent diameter of the Schwarzschild radius is about 40 microarcseconds, or around 10^-10 radians.
This is how small you would have to see to detect those stars whose deflection angle is a radian or more.

The apparent radius of the Einstein ring in radians is approximately the square root of this i.e. 10^-5 radians or about 2 arc seconds, very similar to Einstein’s number, but this is just a coincidence.
Look up the Einstein ring formula in wikipedia and you can do a precise calculation if you want.

For a black hole, light is deflected more and more without limit as a beam approaches the hole. There is no theoretical limit to the angle, but there is a practical limit as the images get closer together.

A sketch of the null geodesics forming
the first two relativistic images on
the primary image side of the optic
axis. The figure on the left
represents the first relativistic
image, which forms when a photon loops
around the black hole once. The figure
on the right represents the second
relativistic image which forms even
closer to the black hole, after
looping around twice. In reality, both
images are very close together on the
lens plane and very close to the black
hole’s photon sphere.

I edited you post by including the picture. The ring demarks the occurrence of multiple images which is the strong gravitational lensing case. Weak gravitational lensing involves elliptical distortion. If you venture to look close to the horizon there are multiple wrappings of these images. The Abel galaxy cluster is hefty enough to produce multiple images and is a strong gravitational lens.
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Lawrence B. CrowellFeb 22 '11 at 14:26

@Lawrence B. Crowell, what is the source of that picture, please? I'm assuming it's an artist's representation.
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finbotMar 4 '11 at 0:30